2023 AMC 10B
All 25 problems, 75 minutes, scored 6 / 1.5 / 0. The answer key and the idea behind each problem are on this page too, folded away until you ask for them.
The problems
Show the answer key, topics and key insights
This gives away all 25 answers and the idea behind each one. Sit the paper first if you mean to.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | C | Ratios, Percents & Averages | Total juice is 3 + 1/3 = 10/3 glasses; shared equally each glass holds 5/6, so each full glass gives up 1/6. | |
| 2 | B | Ratios, Percents & Averages | A 20% discount then 7.5% tax multiplies the price by 0.8 times 1.075 = 0.86, so the budget allows 43/0.86 = 50 dollars. | |
| 3 | D | Circles | The hypotenuse of an inscribed right triangle is a diameter, so the radii are 5/2 and 13/2 and areas scale as (5/13)^2. | |
| 4 | C | Sets, Estimation & Miscellaneous | Convert both dimensions to centimeters first: 6.5 mm = 0.65 cm and 25 m = 2500 cm, then multiply to get 1625 square centimeters. | |
| 5 | A | Linear Equations & Word Problems | Adding 3 to each of n numbers adds 3n to the sum, while tripling each number triples the sum, so S + 3n = 45 = 3S. | |
| 6 | E | Modular Arithmetic | Parities go odd, odd, even and repeat with period 3, so L_n is even exactly when 3 divides n; count multiples of 3 up to 2023. | |
| 7 | B | Transformations & Symmetry | Both squares share a circumcircle; arc AB is 90 degrees and arc AE is 20 degrees, so inscribed angle EAB is half of arc EB = 70 degrees. | |
| 8 | A | Modular Arithmetic | Units digits of powers of 2 and 3 cycle with period 4; exponents 2023 and 2022 leave remainders 3 and 2, giving 8 + 9 = 17. | |
| 9 | B | Number Properties | Consecutive squares differ by (n+1)^2 - n^2 = 2n+1, so the condition 2n+1 <= 2023 means n = 1, 2, ..., 1011. | |
| 10 | C | Paths & Grids | Every domino contains an edge-middle square, so those four squares always work; three squares touch at most 4+3+3 of the 12 dominoes. | |
| 11 | B | Diophantine Equations | Reduce to 2a + 5b + 10c = 80; b is even, so b = 2k gives a = 40 - 5(k+c) with 2 <= k+c <= 7. | |
| 12 | C | Polynomials | P is positive for x > 10 and its sign flips only when crossing a root of odd multiplicity; track the flips at 9, 7, 5, 3, 1. | |
| 13 | B | Absolute Value & Inequalities | By symmetry in both axes, restrict to x, y >= 0 where the region is the diamond |x-1| + |y-1| <= 1 of area 2; multiply by 4. | |
| 14 | C | Diophantine Equations | Add mn to both sides: (m+n)^2 = (mn)^2 + mn, which is strictly between consecutive squares unless mn is 0 or -1. | |
| 15 | C | Number Properties | Pair (2k-1)! with (2k)! = (2k-1)! times 2k: the product is a square times 2 * 4 * ... * 16 = 2^8 * 8!, whose non-square part is 70. | |
| 16 | E | Basic Counting | An upno is a subset of {1,...,9} of size at least 2 and a downno a subset of {0,...,9}; the counts are 2^9 - 10 and 2^10 - 11. | |
| 17 | D | Algebraic Manipulation | The diagonal is sqrt(a^2+b^2+c^2) = sqrt((a+b+c)^2 - 2(ab+bc+ca)); the edge sum and face sum give both pieces without finding a, b, c. | |
| 18 | E | GCD & LCM | Clear denominators: c = 15a + 14b, so c mod 2, 7 depends only on a and c mod 3, 5 only on b; gcd(c,210)=1 iff both gcds are 1. | |
| 19 | B | Geometric Probability | By symmetry fix the direction north; she exits iff y + d > 6, a triangle of area 1/2 inside the (d, y) rectangle of area 6. | |
| 20 | A | Solid Geometry | The four junction points are equally spaced on a great circle, so each semicircle's diameter is a chord of length 2*sqrt(2); four semicircles of radius sqrt(2) total 4*pi*sqrt(2). | |
| 21 | E | Conditional Probability & States | Track only whether all three bin parities agree; each ball flips one parity, giving p_{k+1} = (1 - p_k)/3, which converges to 1/4. | |
| 22 | B | Number Properties | Set k = floor(x); then x = (k^2 + 2)/3 must satisfy k <= x < k + 1, which pins k to 0, 1, 2, 3. | |
| 23 | B | Sequences & Series | The true sum is 221 or 223, so n divides 442 or 446; d >= 2 forces n <= 14, leaving n = 13, d = 2, a = 5. | |
| 24 | E | Coordinate Geometry | The region is the sum of three segments (2,0), (0,1), (-3,4); such a region is a centrally symmetric hexagon whose sides are these segments, each used twice. | |
| 25 | B | Angles & Polygons | Each crease is the perpendicular bisector of a vertex-to-center segment, so the new pentagon has apothem R/2; its scale factor is 1/(2cos36) = 1/phi, area ratio (3-sqrt5)/2. |
Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic tags, difficulty ratings and key insights on this page are original to this site. No problem statements are reproduced here — each links to its own page.